Question

stepwise solution lim(--> infinity) (2x^2-1)^1/2-(2x^2-3x)^1/2

Answer 1

whenever you wee something that is the difference of two square roots, you usually have to multiply by its conjugate

Answer 2

see*

Answer 3

\[\sqrt{2x^2-1}-\sqrt{2x^2-3x}\cdot{\sqrt{2x^2-1}+\sqrt{2x^2-3x}\over\sqrt{2x^2-1}+\sqrt{2x^2-3x}}\]

Answer 4

remember that\[(a+b)(a-b)=a^2-b^2\]

Answer 5

yeah. you're really good with math

Answer 6

thanks, practice is all...

Answer 7

ok, let me work this problem out and see if i get it

Answer 8

I haven't tried it either, I just know how these usually go

Answer 9

but smart people have given me medals, so I bet it's gonna work :)

Answer 10

lol. you're smart too

Answer 11

lol, only when I'm not busy being stupid
'gotta go, good luck!

Answer 12

thank u! bye and take care

Answer 13

actually this still stays \[\LARGE \infty\] I was trying to solve it, I guess I couldn't! just got a circle back to the beginning... L'Hospitals doesn't work either ... wolframa proves it ! http://www.wolframalpha.com/input/?i=lim_%7Bx%5Cto+%5Cinfty%7D+%282x%5E2-1%29%5E%281%2F2%29-%282x%5E2-3x%29%5E%281%2F2%29

Answer 14

i couldn't solve it either. i thought it DNE

Answer 15

DNE= ? :O

Answer 16

does not exist

Answer 17

ahh .. hahaha lol :P These shortcuts you use, I don't get them at all lol. :)

Answer 18

lol. i used that wolfram website too and the answer was definite
http://www.wolframalpha.com/input/?i=limits&a=*C.limits-_*Calculator.dflt-&f2= √%282x%5E2-1%29-√%282x%5E2-3x%29&x=4&y=8&f=Limit.limitfunction_√%282x%5E2-1%29-√%282x%5E2-3x%29&f3=∞&f=Limit.limit_∞&a=*FVarOpt.1-_**-.***Limit.limitvariable--.**Limit.direction--.**Limit.limitvariable2-.*Limit.limit2-.*Limit.direction2---.*--

Answer 19

\[\lim_{x \rightarrow \infty} \left( \sqrt{2x^{2} -1}-\sqrt{2x^2 - 3x} \right)\]multiply by its conjugate\[\lim_{x \rightarrow \infty} \left( \sqrt{2x^{2} -1}-\sqrt{2x^2 - 3x} \right)\left( \frac{\sqrt{2x^2 - 1}+\sqrt{2x^2 - 3x}}{\sqrt{2x^2 - 1}+\sqrt{2x^2 - 3x}} \right)\]\[\lim_{x \rightarrow \infty}\left( \frac{3x - 1}{\sqrt{2x^2 - 1}+\sqrt{2x^2 - 3x}} \right)\left( \frac{\frac{1}{x}}{\frac{1}{x}} \right)\]\[\lim_{x \rightarrow \infty}\left( \frac{3-\frac{1}{x}}{\sqrt{2-\frac{1}{x^2}}+\sqrt{2-\frac{3}{x}}} \right)\]\[\frac{3-0}{\sqrt{2-0}+\sqrt{2-0}}\]\[\frac{3}{2\sqrt{2}}\]

Answer 20

where did the 1/x come from?

Answer 21

I multiply the numerator and denominator by 1/x

Answer 22

oh ok. why?

Answer 23

i think i know why. to cancel out the x from 3x?

Answer 24

i don't understand why 1/x is 0 as x approaches \[\infty\]

Answer 25

\[\lim_{x \rightarrow \infty}\frac{1}{x}=0\]the limit as x approaches infinity of (1/x) is 0, to understand this, try for a really large value for x, you will get results that are nearly 0

Answer 26

oh ok! thank you. Also, when you multiplied by 1/x how did you derive the denominator to be \[\sqrt{2-1/x^2}\]

Answer 27

\[\frac{\sqrt{2x^2 - 1}}{x}=\sqrt{\frac{2x^2 - 1}{x^2}}=\sqrt{2 - \frac{1}{x^2}}\]

Answer 28

speachless... I feel soo stupid! :$

I should've realized the part when we have to divide by x but I only thought about substituting. I guess I got blind totally. Great job @exraven